Question

Find equation of a line passing through the point $(3,1,2)$ and perpendicular to the lines $\overline{r}=(1+\lambda )\overline{i}+(2+2\lambda )\overline{j}+(3+3\lambda )\overline{k}$ and $\overline{r}=(-3\lambda )\overline{i}+\left(2\lambda \right)\overline{j}+\left(5\lambda \right)\overline{k}$.

• A. $\overline{r}=(1.5+\lambda )\overline{i}+(2+\lambda )\overline{j}+(3+\lambda )\overline{k}$
• B. $\overline{r}=(3+2\lambda )\overline{i}+(1-7\lambda )\overline{j}+(2+4\lambda )\overline{k}$
• C. $\overline{r}=2(1.5+\lambda )\overline{i}+(2+\lambda )\overline{j}+(3+\lambda )\overline{k}$
• D. $\overline{r}=2(2+\lambda )\overline{i}+(2+7\lambda )\overline{j}+(3+\lambda )\overline{k}$

Answer 1

The correct option is B $\overline{r}=(3+2\lambda )\overline{i}+(1-7\lambda )\overline{j}+(2+4\lambda )\overline{k}$

Given line is passing through A(3,1,2)

So position vector of line be

$\overrightarrow{a}=3\hat{i}+\hat{j}+2\hat{k}$

So eq of line become

$\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$

line is perpendicular to

$\overrightarrow{r}=(1+\lambda )\hat{i}+(2+2\lambda )\hat{j}+(3+3\lambda )\hat{k}$

$\overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (\hat{i}+2\hat{j}+3\hat{k})$

normal vector of above line

$\overrightarrow{n_1}=\hat{i}+2\hat{j}+3\hat{k}$

line is also perpendicular to

$\overrightarrow{r}=-3\lambda \hat{i}+2\lambda \hat{j}+5\lambda \hat{k}$

$\overrightarrow{r}=\lambda (-3\hat{i}+2\hat{j}+5\hat{k})$

normal vector of above line

$\overrightarrow{n_2}=-3\hat{i}+2\hat{j}+5\hat{k}$

line is perpendicular to both lines so

$\overrightarrow{b}=\overrightarrow{n_1}\times \overrightarrow{n_2}$

$\overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ -3& 2& 5\end{array}\right|$

$\overrightarrow{b}=\hat{i}(10-6)-\hat{j}(5+9)+\hat{k}(2+6)$

$\overrightarrow{b}=4\hat{i}-14\hat{j}+8\hat{k}$

So required line of eq

$\overrightarrow{r}=3\hat{i}+\hat{j}+2\hat{k}+\lambda (4\hat{i}-14\hat{j}+8\hat{k})$

$\overrightarrow{r}=3\hat{i}+\hat{j}+2\hat{k}+2\lambda (2\hat{i}-7\hat{j}+4\hat{k})$

$\overrightarrow{r}=3\hat{i}+\hat{j}+2\hat{k}+\lambda (2\hat{i}-7\hat{j}+4\hat{k})$

$\overrightarrow{r}=(3+2\lambda )\hat{i}+(1-7\lambda )\hat{j}+(2+4\lambda )\hat{k}$

This is required eq